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Either paper 2 marks
Write down all the integers that satisfy the inequality
\(-4 ≤ m < 0\)
Show answer
\(-4,\ -3,\ -2,\ -1\)Lower end: \(≤\) means -4 is kept (≤ includes the end value).
Upper end: \(<\) means 0 is dropped (< excludes the end value).
So the integers run from -4 to -1, counting up through every whole number in between, passing through zero where the range crosses it:
\(-4,\ -3,\ -2,\ -1\)
- B1both end integers handled correctly, so the list runs from \(-4\) to \(-1\)
- B1every integer from \(-4\) to \(-1\) listed, none missing
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Either paper 2 marks
Solve the inequality \(8x - 1 > 23\)
Show answer
\(x > 3\)Add 1 to both sides:
\(8x > 24\)
Divide both sides by \(8\). The number is positive, so the inequality sign does not change:
\(x > 3\)
- M1subtracting or adding to get \(8x > 24\)
- A1dividing by \(8\) to reach \(x > 3\)
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Either paper 2 marks
Solve the inequality \(4(x - 9) > -40\)
Show answer
\(x > -1\)Divide both sides by \(4\) to remove the bracket. The number is positive, so the inequality sign does not change:
\(x - 9 > -10\)
Add 9 to both sides:
\(x > -1\)
(Expanding first gives \(4x - 36 > -40\), which leads to the same answer.)
- M1removes the bracket: divides both sides by \(4\) to get \(x - 9 > -10\), or expands to \(4x - 36 > -40\)
- A1solves to \(x > -1\), sign kept
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Either paper 4 marks
Two inequalities in \(m\) are set out in parts (a) and (b).
(a)Solve \(4 ≤ 4m < 20\)
(b)Solve \(4 < 4m ≤ 20\)
(c)\(m\) is an integer.
How are the integer solutions to part (a) and to part (b) different?
Show answer
(a) \(1 ≤ m < 5\)
(b) \(1 < m ≤ 5\)
(c) Only (b) has \(5\) as a solution, and only (a) has \(1\) as a solutionDivide all three parts by \(4\). The number is positive, so neither sign changes:
\(1 ≤ m < 5\)
Divide all three parts by \(4\):
\(1 < m ≤ 5\)
Integers for (a): \(1,\ 2,\ 3,\ 4\)
Integers for (b): \(2,\ 3,\ 4,\ 5\)
The two lists agree except at the ends. (b) includes \(5\) and the other list does not. (a) includes \(1\) and the other list does not.
- (a) B1dividing all three parts by \(4\), with the signs kept
- (b) B1dividing all three parts by \(4\), with the signs kept
- (c) M1the integers found for (a) \(1,\ 2,\ 3,\ 4\) and for (b) \(2,\ 3,\ 4,\ 5\)
- (c) A1(b) alone has \(5\) and (a) alone has \(1\)
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Either paper 2 marks
The solution of the inequality
\(5x + b > 43\)
is \(x > 7\).
Work out the value of \(b\).
Show answer
\(8\)The inequality turns into an equality at the end of the solution set, so \(x = 7\) is where the two sides are equal.
\(5 \times 7 + b = 43\)
\(35 + b = 43\)
\(b = 43 - 35 = 8\)
Dividing by the positive number 5 never flips the sign, which is why the solution keeps the \(>\) it started with.
- M1substituting \(x = 7\) to get \(5 \times 7 + b = 43\)
- A1\(b = 8\)
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Either paper 2 marks
Erin is asked to show the inequality \(1 \leq x < 5\) on a number line.
Here is her attempt.
State the two mistakes she has made.
Show answer
At the left end the mistake is the circle: it is drawn as an open circle at \(x = 1\), but it should be a filled-in circle at \(x = 1\). At the right end the mistake is the circle: it is drawn as a filled-in circle at \(x = 5\), but it should be an open circle at \(x = 5\).The stated inequality is \(1 \leq x < 5\), so the circle at \(1\) should be a filled-in circle and the circle at \(5\) should be an open circle.
Compare that with what is actually drawn.
At the left end the diagram shows an open circle at \(x = 1\) — the mistake there is the circle.
At the right end the diagram shows a filled-in circle at \(x = 5\) — the mistake there is the circle.
- B1left end mistake identified: the circle, should be a filled-in circle at \(x = 1\)
- B1right end mistake identified: the circle, should be an open circle at \(x = 5\)
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Either paper 1 mark
\(a < 0\) and \(b > 0\).
Which of the statements below must always be true? Write it down.
A: \(ab > 0\)
B: \(a^2 < 0\)
C: \(a - b > 0\)
D: \(b - a > 0\)
Show answer
b - a > 0\(a < 0\) means \(-a > 0\), so \(b - a\) is positive \(+\) positive: \(b - a > 0\) is always true.
\(ab\) is negative × positive, so \(ab < 0\) is always true and \(ab > 0\) is always false.
\(a^2\) is a square, so it can never be negative: \(a^2 < 0\) is impossible.
\(a - b\) is negative \(-\) positive, which is always negative, so \(a - b > 0\) is false.
The statement that must always be true is \(b - a > 0\).
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Either paper 1 mark
\(6 < \sqrt{ x } < 7\)
Circle the possible value of \(x\).
\(2.55,\ 6.5,\ 13,\ 42\)
Show answer
\(42\)To undo \(\sqrt{ x }\), squared every part of the inequality:
\(6^2 < x < 7^2\)
\(36 < x < 49\)
Only \(42\) lies between \(36\) and \(49\).
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Either paper 2 marks
Solve \(8x - 13 \le 5x + 2\)
Show answer
\(x \le 5\)Collect the \(x\) terms on the left by subtracting \(5x\) from both sides:
\(3x - 13 \le 2\)
Add 13 to both sides to leave the \(x\) term alone:
\(3x \le 15\)
Divide both sides by \(3\). The number is positive, so the inequality sign does not change:
\(x \le 5\)
- M1collecting the \(x\) terms and the numbers to get \(3x \le 15\)
- A1dividing by \(3\) to reach \(x \le 5\)
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Either paper 3 marks
Solve \(1 ≤ 2x - 1 < 5\)
Show answer
\(1 ≤ x < 3\)The same operation must be done to all three parts of the inequality.
Add 1 to all three parts:
\(2 ≤ 2x < 6\)
Divide all three parts by \(2\). The number is positive, so neither inequality sign changes:
\(1 ≤ x < 3\)
- M1Add 1 to all three parts: \(2 ≤ 2x < 6\)
- M1divide all three parts by \(2\): \(1 ≤ x < 3\)
- A1ends \(1\) and \(3\) kept in order, with signs \(≤\) and \(<\) unchanged